What is the length of GF?

Let the radius of the semicircle be $r$ and the area of the region be $A_{\red{\text{red}}}$ . Then the area of the white part of the semicircle is $A_{\text{white}} = \dfrac {\pi r^2}2 - A_{\red{\text{red}}}$ , and the area of the yellow region $A_{\color{#CEBB00}{\text{yellow}}} = \dfrac \pi 4 - A_{\text{white}} = \dfrac \pi 4 - \dfrac {\pi r^2}2 + A_{\red{\text{red}}}$ . Now we have:

$\begin{aligned} A_{\color{#CEBB00}{\text{yellow}}} & = A_{\red{\text{red}}} \\ \frac \pi 4 - \frac {\pi r^2}2 + A_{\red{\text{red}}} & = A_{\red{\text{red}}} \\ \frac \pi 4 - \frac {\pi r^2}2 & = 0 \\ \frac {\pi r^2}2 & = \frac \pi 4 \\ \implies r & = \frac 1{\sqrt 2} \\ GF^2 & = (2r)^2 = \boxed 2 \end{aligned}$

Let the breadth of the rectangle be $b$ and the area of the red region be $A$ . Then $A=\dfrac{π}{4}-\left (\dfrac{π(b+1)^2}{8}-A\right )\implies b=\sqrt 2-1\implies |\overline {GF}|=\sqrt 2\implies |\overline {GF}|^2=\boxed 2$ .